How to get an equation from a “ball is thrown” word problem?
0
$begingroup$
"A ball is thrown upward from the roof of a building 60m tall. The ball reaches a height of 80m above the ground after 2 s and hits the ground 6 s after being thrown." How do I determine the equation without using physics(constants related to physics) ideas?
It should be in the form y= ax^2+ bx + c, right? I got 60 for the y-intercept so y= ax^2+ bx + 60 The ball had the height of 80 meters at 2 seconds. It dropped to the ground in 4 seconds. So 80/4=b b=20? I found a and got -5x^2 + 20x + 60 but do my steps show I understand the problem?
functions word-problem
asked Oct 1 '16 at 2:44
HamzeHamze
195
$endgroup$
|
show 3 more comments
0
$begingroup$
"A ball is thrown upward from the roof of a building 60m tall. The ball reaches a height of 80m above the ground after 2 s and hits the ground 6 s after being thrown." How do I determine the equation without using physics(constants related to physics) ideas?
It should be in the form y= ax^2+ bx + c, right? I got 60 for the y-intercept so y= ax^2+ bx + 60 The ball had the height of 80 meters at 2 seconds. It dropped to the ground in 4 seconds. So 80/4=b b=20? I found a and got -5x^2 + 20x + 60 but do my steps show I understand the problem?
functions word-problem
asked Oct 1 '16 at 2:44
HamzeHamze
195
$endgroup$
1
$begingroup$
How can one answer a physics question without using physics? And what is the question?
$endgroup$
– Mark Viola
Oct 1 '16 at 2:47
$begingroup$
@Dr.MV: when there is too much information supplied, so there is enough to measure $g$. We still need the assumption that $g$ is constant.
$endgroup$
– Ross Millikan
Oct 1 '16 at 2:53
$begingroup$
Please define your variables and show more steps. You appear to start with $x$ for time in seconds, but at the end $t$ appears. How did you get to $80/4=b?$ There is also a typo of $6$ for $60$ in the $y$ intercept. I suspect you understand the problem and $a$ is not the canonical $9.81$, but you haven't shown enough steps to show you understand it.
$endgroup$
– Ross Millikan
Oct 1 '16 at 2:57
$begingroup$
@RossMillikan Ross, what is the question that the OP is trying to answer? And how does one answer a physics question without physics? That is, what defined the trajectory (path) of the object.
$endgroup$
– Mark Viola
Oct 1 '16 at 2:59
$begingroup$
@Dr.MV: once you accept the trajectory is quadratic, which to my mind is certainly physics, you can evaluate $a$ and $b$ from the data given. You don't have to accept that $a=-9.81$. I find a different value for $a$
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:08
|
show 3 more comments
0
0
0
$begingroup$
"A ball is thrown upward from the roof of a building 60m tall. The ball reaches a height of 80m above the ground after 2 s and hits the ground 6 s after being thrown." How do I determine the equation without using physics(constants related to physics) ideas?
It should be in the form y= ax^2+ bx + c, right? I got 60 for the y-intercept so y= ax^2+ bx + 60 The ball had the height of 80 meters at 2 seconds. It dropped to the ground in 4 seconds. So 80/4=b b=20? I found a and got -5x^2 + 20x + 60 but do my steps show I understand the problem?
functions word-problem
asked Oct 1 '16 at 2:44
HamzeHamze
195
$endgroup$
"A ball is thrown upward from the roof of a building 60m tall. The ball reaches a height of 80m above the ground after 2 s and hits the ground 6 s after being thrown." How do I determine the equation without using physics(constants related to physics) ideas?
It should be in the form y= ax^2+ bx + c, right? I got 60 for the y-intercept so y= ax^2+ bx + 60 The ball had the height of 80 meters at 2 seconds. It dropped to the ground in 4 seconds. So 80/4=b b=20? I found a and got -5x^2 + 20x + 60 but do my steps show I understand the problem?
functions word-problem
functions word-problem
asked Oct 1 '16 at 2:44
HamzeHamze
195
asked Oct 1 '16 at 2:44
HamzeHamze
195
edited Oct 1 '16 at 3:16
Hamze
asked Oct 1 '16 at 2:44
HamzeHamze
195
asked Oct 1 '16 at 2:44
HamzeHamze
195
asked Oct 1 '16 at 2:44
HamzeHamze
195
195
1
$begingroup$
How can one answer a physics question without using physics? And what is the question?
$endgroup$
– Mark Viola
Oct 1 '16 at 2:47
$begingroup$
@Dr.MV: when there is too much information supplied, so there is enough to measure $g$. We still need the assumption that $g$ is constant.
$endgroup$
– Ross Millikan
Oct 1 '16 at 2:53
$begingroup$
Please define your variables and show more steps. You appear to start with $x$ for time in seconds, but at the end $t$ appears. How did you get to $80/4=b?$ There is also a typo of $6$ for $60$ in the $y$ intercept. I suspect you understand the problem and $a$ is not the canonical $9.81$, but you haven't shown enough steps to show you understand it.
$endgroup$
– Ross Millikan
Oct 1 '16 at 2:57
$begingroup$
@RossMillikan Ross, what is the question that the OP is trying to answer? And how does one answer a physics question without physics? That is, what defined the trajectory (path) of the object.
$endgroup$
– Mark Viola
Oct 1 '16 at 2:59
$begingroup$
@Dr.MV: once you accept the trajectory is quadratic, which to my mind is certainly physics, you can evaluate $a$ and $b$ from the data given. You don't have to accept that $a=-9.81$. I find a different value for $a$
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:08
|
show 3 more comments
1
$begingroup$
How can one answer a physics question without using physics? And what is the question?
$endgroup$
– Mark Viola
Oct 1 '16 at 2:47
$begingroup$
@Dr.MV: when there is too much information supplied, so there is enough to measure $g$. We still need the assumption that $g$ is constant.
$endgroup$
– Ross Millikan
Oct 1 '16 at 2:53
$begingroup$
Please define your variables and show more steps. You appear to start with $x$ for time in seconds, but at the end $t$ appears. How did you get to $80/4=b?$ There is also a typo of $6$ for $60$ in the $y$ intercept. I suspect you understand the problem and $a$ is not the canonical $9.81$, but you haven't shown enough steps to show you understand it.
$endgroup$
– Ross Millikan
Oct 1 '16 at 2:57
$begingroup$
@RossMillikan Ross, what is the question that the OP is trying to answer? And how does one answer a physics question without physics? That is, what defined the trajectory (path) of the object.
$endgroup$
– Mark Viola
Oct 1 '16 at 2:59
$begingroup$
@Dr.MV: once you accept the trajectory is quadratic, which to my mind is certainly physics, you can evaluate $a$ and $b$ from the data given. You don't have to accept that $a=-9.81$. I find a different value for $a$
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:08
1
1
$begingroup$
How can one answer a physics question without using physics? And what is the question?
$endgroup$
– Mark Viola
Oct 1 '16 at 2:47
$begingroup$
How can one answer a physics question without using physics? And what is the question?
$endgroup$
– Mark Viola
Oct 1 '16 at 2:47
$begingroup$
@Dr.MV: when there is too much information supplied, so there is enough to measure $g$. We still need the assumption that $g$ is constant.
$endgroup$
– Ross Millikan
Oct 1 '16 at 2:53
$begingroup$
@Dr.MV: when there is too much information supplied, so there is enough to measure $g$. We still need the assumption that $g$ is constant.
$endgroup$
– Ross Millikan
Oct 1 '16 at 2:53
$begingroup$
Please define your variables and show more steps. You appear to start with $x$ for time in seconds, but at the end $t$ appears. How did you get to $80/4=b?$ There is also a typo of $6$ for $60$ in the $y$ intercept. I suspect you understand the problem and $a$ is not the canonical $9.81$, but you haven't shown enough steps to show you understand it.
$endgroup$
– Ross Millikan
Oct 1 '16 at 2:57
$begingroup$
Please define your variables and show more steps. You appear to start with $x$ for time in seconds, but at the end $t$ appears. How did you get to $80/4=b?$ There is also a typo of $6$ for $60$ in the $y$ intercept. I suspect you understand the problem and $a$ is not the canonical $9.81$, but you haven't shown enough steps to show you understand it.
$endgroup$
– Ross Millikan
Oct 1 '16 at 2:57
$begingroup$
@RossMillikan Ross, what is the question that the OP is trying to answer? And how does one answer a physics question without physics? That is, what defined the trajectory (path) of the object.
$endgroup$
– Mark Viola
Oct 1 '16 at 2:59
$begingroup$
@RossMillikan Ross, what is the question that the OP is trying to answer? And how does one answer a physics question without physics? That is, what defined the trajectory (path) of the object.
$endgroup$
– Mark Viola
Oct 1 '16 at 2:59
$begingroup$
@Dr.MV: once you accept the trajectory is quadratic, which to my mind is certainly physics, you can evaluate $a$ and $b$ from the data given. You don't have to accept that $a=-9.81$. I find a different value for $a$
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:08
$begingroup$
@Dr.MV: once you accept the trajectory is quadratic, which to my mind is certainly physics, you can evaluate $a$ and $b$ from the data given. You don't have to accept that $a=-9.81$. I find a different value for $a$
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:08
|
show 3 more comments
1 Answer
1
active
oldest
votes
0
$begingroup$
Physics says two things. First, that the acceleration from gravity is constant, and second, that the acceleration is $-9.81 m/s^2$. You have accepted the first, which results in $y=ax^2+bx+c$ (assuming $x$ is time) but rejected the second. That is fine, you have enough data to evaluate all the constants, as you have three points on the parabola and need three constants, $a,b,c$. As you say, $c=60$ is immediate.
answered Oct 1 '16 at 3:17
Ross MillikanRoss Millikan
293k23197371
$endgroup$
1
$begingroup$
We know the equation is y=ax^2+ bx +60. I'd put in the point (2,80) to get 80=a(2)^2 + b(2) +60. Put in the point (6,0) to get 0=a(6)^2 + b(6) +60. Now I have 2 equations with 2 unknowns. I used all 3 points to find all 3 constants by solving for a, and then b. Does this work?
$endgroup$
– Hamze
Oct 1 '16 at 3:23
$begingroup$
Yes, that works fine. It turns out $b=20$, but your logic in the question does not justify that.
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:30
$begingroup$
I don't understand why my method works, actually. Did I find the point of intersection? Could I have used any other 2 points?
$endgroup$
– Hamze
Oct 1 '16 at 3:35
$begingroup$
You were lucky. It turns out $(2,80)$ is the peak, but that is not given. You can write it as $y=-5(x-2)^2+80$. If the ground were at the launch altitude the equation would be $y=-5(x-2)^2$ and it would have hit the ground at $4$ seconds. You would have found $b=20/2=10$ because the peak was $20$ and it hit the ground $2$ seconds later.
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:49
1
$begingroup$
You have three points on the parabola, if you add in $(0,60)$. That gives you three equations in the three unknowns $a,b,c$ You are just solving the three simultaneous equations to find them. As you noted in the question, the fact that you have a value at $x=0$ makes finding $c$ immediate.
$endgroup$
– Ross Millikan
Oct 1 '16 at 4:03
|
show 1 more comment
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1 Answer
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oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$begingroup$
Physics says two things. First, that the acceleration from gravity is constant, and second, that the acceleration is $-9.81 m/s^2$. You have accepted the first, which results in $y=ax^2+bx+c$ (assuming $x$ is time) but rejected the second. That is fine, you have enough data to evaluate all the constants, as you have three points on the parabola and need three constants, $a,b,c$. As you say, $c=60$ is immediate.
answered Oct 1 '16 at 3:17
Ross MillikanRoss Millikan
293k23197371
$endgroup$
1
$begingroup$
We know the equation is y=ax^2+ bx +60. I'd put in the point (2,80) to get 80=a(2)^2 + b(2) +60. Put in the point (6,0) to get 0=a(6)^2 + b(6) +60. Now I have 2 equations with 2 unknowns. I used all 3 points to find all 3 constants by solving for a, and then b. Does this work?
$endgroup$
– Hamze
Oct 1 '16 at 3:23
$begingroup$
Yes, that works fine. It turns out $b=20$, but your logic in the question does not justify that.
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:30
$begingroup$
I don't understand why my method works, actually. Did I find the point of intersection? Could I have used any other 2 points?
$endgroup$
– Hamze
Oct 1 '16 at 3:35
$begingroup$
You were lucky. It turns out $(2,80)$ is the peak, but that is not given. You can write it as $y=-5(x-2)^2+80$. If the ground were at the launch altitude the equation would be $y=-5(x-2)^2$ and it would have hit the ground at $4$ seconds. You would have found $b=20/2=10$ because the peak was $20$ and it hit the ground $2$ seconds later.
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:49
1
$begingroup$
You have three points on the parabola, if you add in $(0,60)$. That gives you three equations in the three unknowns $a,b,c$ You are just solving the three simultaneous equations to find them. As you noted in the question, the fact that you have a value at $x=0$ makes finding $c$ immediate.
$endgroup$
– Ross Millikan
Oct 1 '16 at 4:03
|
show 1 more comment
0
$begingroup$
Physics says two things. First, that the acceleration from gravity is constant, and second, that the acceleration is $-9.81 m/s^2$. You have accepted the first, which results in $y=ax^2+bx+c$ (assuming $x$ is time) but rejected the second. That is fine, you have enough data to evaluate all the constants, as you have three points on the parabola and need three constants, $a,b,c$. As you say, $c=60$ is immediate.
answered Oct 1 '16 at 3:17
Ross MillikanRoss Millikan
293k23197371
$endgroup$
1
$begingroup$
We know the equation is y=ax^2+ bx +60. I'd put in the point (2,80) to get 80=a(2)^2 + b(2) +60. Put in the point (6,0) to get 0=a(6)^2 + b(6) +60. Now I have 2 equations with 2 unknowns. I used all 3 points to find all 3 constants by solving for a, and then b. Does this work?
$endgroup$
– Hamze
Oct 1 '16 at 3:23
$begingroup$
Yes, that works fine. It turns out $b=20$, but your logic in the question does not justify that.
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:30
$begingroup$
I don't understand why my method works, actually. Did I find the point of intersection? Could I have used any other 2 points?
$endgroup$
– Hamze
Oct 1 '16 at 3:35
$begingroup$
You were lucky. It turns out $(2,80)$ is the peak, but that is not given. You can write it as $y=-5(x-2)^2+80$. If the ground were at the launch altitude the equation would be $y=-5(x-2)^2$ and it would have hit the ground at $4$ seconds. You would have found $b=20/2=10$ because the peak was $20$ and it hit the ground $2$ seconds later.
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:49
1
$begingroup$
You have three points on the parabola, if you add in $(0,60)$. That gives you three equations in the three unknowns $a,b,c$ You are just solving the three simultaneous equations to find them. As you noted in the question, the fact that you have a value at $x=0$ makes finding $c$ immediate.
$endgroup$
– Ross Millikan
Oct 1 '16 at 4:03
|
show 1 more comment
0
0
0
$begingroup$
Physics says two things. First, that the acceleration from gravity is constant, and second, that the acceleration is $-9.81 m/s^2$. You have accepted the first, which results in $y=ax^2+bx+c$ (assuming $x$ is time) but rejected the second. That is fine, you have enough data to evaluate all the constants, as you have three points on the parabola and need three constants, $a,b,c$. As you say, $c=60$ is immediate.
answered Oct 1 '16 at 3:17
Ross MillikanRoss Millikan
293k23197371
$endgroup$
Physics says two things. First, that the acceleration from gravity is constant, and second, that the acceleration is $-9.81 m/s^2$. You have accepted the first, which results in $y=ax^2+bx+c$ (assuming $x$ is time) but rejected the second. That is fine, you have enough data to evaluate all the constants, as you have three points on the parabola and need three constants, $a,b,c$. As you say, $c=60$ is immediate.
answered Oct 1 '16 at 3:17
Ross MillikanRoss Millikan
293k23197371
edited Oct 1 '16 at 3:31
answered Oct 1 '16 at 3:17
Ross MillikanRoss Millikan
293k23197371
answered Oct 1 '16 at 3:17
Ross MillikanRoss Millikan
293k23197371
answered Oct 1 '16 at 3:17
Ross MillikanRoss Millikan
293k23197371
293k23197371
1
$begingroup$
We know the equation is y=ax^2+ bx +60. I'd put in the point (2,80) to get 80=a(2)^2 + b(2) +60. Put in the point (6,0) to get 0=a(6)^2 + b(6) +60. Now I have 2 equations with 2 unknowns. I used all 3 points to find all 3 constants by solving for a, and then b. Does this work?
$endgroup$
– Hamze
Oct 1 '16 at 3:23
$begingroup$
Yes, that works fine. It turns out $b=20$, but your logic in the question does not justify that.
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:30
$begingroup$
I don't understand why my method works, actually. Did I find the point of intersection? Could I have used any other 2 points?
$endgroup$
– Hamze
Oct 1 '16 at 3:35
$begingroup$
You were lucky. It turns out $(2,80)$ is the peak, but that is not given. You can write it as $y=-5(x-2)^2+80$. If the ground were at the launch altitude the equation would be $y=-5(x-2)^2$ and it would have hit the ground at $4$ seconds. You would have found $b=20/2=10$ because the peak was $20$ and it hit the ground $2$ seconds later.
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:49
1
$begingroup$
You have three points on the parabola, if you add in $(0,60)$. That gives you three equations in the three unknowns $a,b,c$ You are just solving the three simultaneous equations to find them. As you noted in the question, the fact that you have a value at $x=0$ makes finding $c$ immediate.
$endgroup$
– Ross Millikan
Oct 1 '16 at 4:03
|
show 1 more comment
1
$begingroup$
We know the equation is y=ax^2+ bx +60. I'd put in the point (2,80) to get 80=a(2)^2 + b(2) +60. Put in the point (6,0) to get 0=a(6)^2 + b(6) +60. Now I have 2 equations with 2 unknowns. I used all 3 points to find all 3 constants by solving for a, and then b. Does this work?
$endgroup$
– Hamze
Oct 1 '16 at 3:23
$begingroup$
Yes, that works fine. It turns out $b=20$, but your logic in the question does not justify that.
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:30
$begingroup$
I don't understand why my method works, actually. Did I find the point of intersection? Could I have used any other 2 points?
$endgroup$
– Hamze
Oct 1 '16 at 3:35
$begingroup$
You were lucky. It turns out $(2,80)$ is the peak, but that is not given. You can write it as $y=-5(x-2)^2+80$. If the ground were at the launch altitude the equation would be $y=-5(x-2)^2$ and it would have hit the ground at $4$ seconds. You would have found $b=20/2=10$ because the peak was $20$ and it hit the ground $2$ seconds later.
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:49
1
$begingroup$
You have three points on the parabola, if you add in $(0,60)$. That gives you three equations in the three unknowns $a,b,c$ You are just solving the three simultaneous equations to find them. As you noted in the question, the fact that you have a value at $x=0$ makes finding $c$ immediate.
$endgroup$
– Ross Millikan
Oct 1 '16 at 4:03
1
1
$begingroup$
We know the equation is y=ax^2+ bx +60. I'd put in the point (2,80) to get 80=a(2)^2 + b(2) +60. Put in the point (6,0) to get 0=a(6)^2 + b(6) +60. Now I have 2 equations with 2 unknowns. I used all 3 points to find all 3 constants by solving for a, and then b. Does this work?
$endgroup$
– Hamze
Oct 1 '16 at 3:23
$begingroup$
We know the equation is y=ax^2+ bx +60. I'd put in the point (2,80) to get 80=a(2)^2 + b(2) +60. Put in the point (6,0) to get 0=a(6)^2 + b(6) +60. Now I have 2 equations with 2 unknowns. I used all 3 points to find all 3 constants by solving for a, and then b. Does this work?
$endgroup$
– Hamze
Oct 1 '16 at 3:23
$begingroup$
Yes, that works fine. It turns out $b=20$, but your logic in the question does not justify that.
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:30
$begingroup$
Yes, that works fine. It turns out $b=20$, but your logic in the question does not justify that.
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:30
$begingroup$
I don't understand why my method works, actually. Did I find the point of intersection? Could I have used any other 2 points?
$endgroup$
– Hamze
Oct 1 '16 at 3:35
$begingroup$
I don't understand why my method works, actually. Did I find the point of intersection? Could I have used any other 2 points?
$endgroup$
– Hamze
Oct 1 '16 at 3:35
$begingroup$
You were lucky. It turns out $(2,80)$ is the peak, but that is not given. You can write it as $y=-5(x-2)^2+80$. If the ground were at the launch altitude the equation would be $y=-5(x-2)^2$ and it would have hit the ground at $4$ seconds. You would have found $b=20/2=10$ because the peak was $20$ and it hit the ground $2$ seconds later.
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:49
$begingroup$
You were lucky. It turns out $(2,80)$ is the peak, but that is not given. You can write it as $y=-5(x-2)^2+80$. If the ground were at the launch altitude the equation would be $y=-5(x-2)^2$ and it would have hit the ground at $4$ seconds. You would have found $b=20/2=10$ because the peak was $20$ and it hit the ground $2$ seconds later.
$endgroup$
– Ross Millikan
Oct 1 '16 at 3:49
1
1
$begingroup$
You have three points on the parabola, if you add in $(0,60)$. That gives you three equations in the three unknowns $a,b,c$ You are just solving the three simultaneous equations to find them. As you noted in the question, the fact that you have a value at $x=0$ makes finding $c$ immediate.
$endgroup$
– Ross Millikan
Oct 1 '16 at 4:03
$begingroup$
You have three points on the parabola, if you add in $(0,60)$. That gives you three equations in the three unknowns $a,b,c$ You are just solving the three simultaneous equations to find them. As you noted in the question, the fact that you have a value at $x=0$ makes finding $c$ immediate.
$endgroup$
– Ross Millikan
Oct 1 '16 at 4:03
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1
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How can one answer a physics question without using physics? And what is the question?
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– Mark Viola
Oct 1 '16 at 2:47
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@Dr.MV: when there is too much information supplied, so there is enough to measure $g$. We still need the assumption that $g$ is constant.
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– Ross Millikan
Oct 1 '16 at 2:53
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Please define your variables and show more steps. You appear to start with $x$ for time in seconds, but at the end $t$ appears. How did you get to $80/4=b?$ There is also a typo of $6$ for $60$ in the $y$ intercept. I suspect you understand the problem and $a$ is not the canonical $9.81$, but you haven't shown enough steps to show you understand it.
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– Ross Millikan
Oct 1 '16 at 2:57
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@RossMillikan Ross, what is the question that the OP is trying to answer? And how does one answer a physics question without physics? That is, what defined the trajectory (path) of the object.
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– Mark Viola
Oct 1 '16 at 2:59
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@Dr.MV: once you accept the trajectory is quadratic, which to my mind is certainly physics, you can evaluate $a$ and $b$ from the data given. You don't have to accept that $a=-9.81$. I find a different value for $a$
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– Ross Millikan
Oct 1 '16 at 3:08